Problem: Simplify the following expression: $y = \dfrac{3x^2+4x- 15}{x + 3}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(-15)} &=& -45 \\ {a} + {b} &=& &=& {4} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-45$ and add them together. Remember, since $-45$ is negative, one of the factors must be negative. The factors that add up to ${4}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-5}$ and ${b}$ is ${9}$ $ \begin{eqnarray} {ab} &=& ({-5})({9}) &=& -45 \\ {a} + {b} &=& {-5} + {9} &=& 4 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({3}x^2 {-5}x) + ({9}x {-15}) $ Factor out the common factors: $ x(3x - 5) + 3(3x - 5)$ Now factor out $(3x - 5)$ $ (3x - 5)(x + 3)$ The original expression can therefore be written: $ \dfrac{(3x - 5)(x + 3)}{x + 3}$ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ This leaves us with $3x - 5; x \neq -3$.